Let \(X\)  be a tetravalent half-arc-transitive graph of order \(12p\)  where p is a prime. Let \(A:=\rm {Aut(X)}\) . Since the stabilizer \(A_{v}\)  of \(v \in V(X)\)  is a 2-group, we have \(|A|= 2^{m+2}.3.p\) , for some nonegative integer \(m\) .
Suppose to the contrary that \(A\)  has no normal Sylow p-subgroups. Let \(N\)  be a minimal normal subgroup of \(A\) . We claim that \(N\)  is solvable. Otherwise, by Proposition REF  and since \(p > 53\) , we get a contradiction. Then \(N\)  is solvable and hence it is an elementary abelian 2-,3- or p-group.
Case I: \(N\)  is a 2-group.
Let \(X_{N}\)  be the quotient graph of \(X\)  corresponding to the orbits of \(N\)  on \(V(X)\) . Then \(|V(X_{N})|= 6p\)  or \(3p\) .
Subcase 1: \(|V(X_{N})|= 6p\) .
Since \(X\)  is edge-transitive, \(X_{N}\)  has valency 2 or 4. Suppose that \(X_{N}\)  has valency 2. Then \(X\cong C_{6p}[2K_{1}]\) , which is arc-transitive. It is a contradiction. Assume now that \(X_{N}\)  has valency 4. If \(X_{N}\)  is half-arc-transitive then by [[1]}, Theorem 4.1], \(|\rm {Aut(X_{N})}|= 2^{2}.3.p\) . Let \(K\)  be the kernel of \(A\)  acting on \(V(X_{N})\) . Since \(K\)  fixes each orbit of \(N\) , the stabilizer \(K_{v}=1\)  for any \(v \in V(X)\) . Then \(|N|=|K|\) . On the other hand \(A/K \leqslant \rm {Aut(X_{N})}\) . Since \(A/K\)  acts transitively on \(V(X_{N})\)  and \(E(X_{N})\) , \(|A|=24p\) . Then \(1+np\)  \(|\)  24. Since \(p > 53\)  then \(P \trianglelefteq A\) , a contradiction.
Now, suppose that \(X_{N}\)  is arc-transitive. Let \(X_{N}\)  has valency 4. By Proposition REF , if \(X_{N}\)  is arc-regular then \(|\rm {Aut(X_{N})}|= 24p\) . By lemma REF , \(A/K\)  is half-arc-transitive and hence \(|A|=24p\) . Then \(P \trianglelefteq A\)  because \(p > 53\)  . It is a contradiction. If \(X_{N}\)  not be arc-regular then by Proposition REF , \(Y=X_{N}\cong C_{3p}[2K_{1}]\)  and \(B= \rm {Aut(Y)}\) . \(|B|= 2^{3p+1}.3.p\) .
Assume that \(M\)  is a minimal normal subgroup of \(B\) . By the same argument as in the first paragraph, \(M\)  is solvable and hence it is an elementary abelian 2-,3- or p-group. First, assume that \(M\)  is a 2-group and \(Y_{M}\)  is the quotient graph of \(Y\)  corresponding to the orbits of \(M\)  on \(V(Y)\) . The quotient graph \(Y_{M}\)  has order \(3p\)  and valency 2 or 4. If \(Y_{M}\)  has valency 4 then \(M_{v}=1\)  for \(v \in V(Y)\) . Assume that \(K_{1}\)  be the kernel of \(B\)  acting on \(V(Y_{M})\) . Hence \(|K_{1}|=|M|\) . Thus \(B/K_{1} \leqslant \rm {Aut(Y_{M})}\) . It is a contradiction because \(\rm {|Aut(Y_{M})}|= 12p\)  by [[2]}, Theorem 5]. If \(Y_{M}\)  has valency 2 then \(Y_{M} \cong C_{3p}\)  and \(\rm {Aut(Y_{M})} \cong D_{6p}\) . Since \(|K_{1}| \le 2\) , we have \(|B| \le 12p\) . We get a contradiction because \(p > 53\) . Now, suppose that \(M\)  be a 3-group. Then \(|V(Y_{M})|= 2p\) . Since \(M_{v}=1\)  for \(v \in V(Y)\)  by using [[3]}, Theorem 1.1(4)], \(Y_{M}\)  has valency 4. By [[4]}, Table 1], \(Y_{M} \cong G(2,p,r)\)  or \(G(2p,r)\) . Then \(|K_{1}|=|M|\)  and hence \(B/K_{1} \leqslant \rm {Aut(Y_{M})}\) . It is a contradiction because \(|\rm {Aut(Y_{M})}|= 2^{p+1}.p\)  or \(8p\)  and \(p > 53\) . Let \(M\)  be a p-group. Then \(|Y_{M}|= 6\) . Since \(M_{v} \le M\)  we have \(|M_{v}|=1\) . By [[3]}, Theorem 1.1(4)], \(Y_{M}\)  has valency 4. By [6]}, \(|\rm {Aut(Y_{M})}|= 48\) . Hence \(B/K_{1} \leqslant \rm {Aut(Y_{M})}\) . It is a contradiction.
Subcase 2: \(|V(X_{N})|= 3p\) .
Let \(|V(X_{N})|= 3p\)  and \(X_{N}\)  has valency 2. Then \(X\cong C_{3p}[2K_{1}]\) . This leads to a contradiction. If \(X_{N}\)  has valency 4 and it is half-arc-transitive then by [[7]}, Theorem 2.5], \(|\rm {Aut(X_{N})}|= 6p\) . Since \(X_{N}\)  is an edge-transitive graph, \(6p~|~|A/K|~|6p\) . Then \(|A|=24p\)  and hence \(P \trianglelefteq A\) . It is a contradiction. Suppose now that \(X_{N}\)  is arc-transitive. By [[2]}, Theorem 5], \(|\rm {Aut(X_{N})}|= 12p\) . Then with the same arguments as before, a contradiction can be obtained.
Case II: \(N\)  is 3-group.
If \(|V(X_{N})|= 4p\)  and \(X_{N}\)  has valency 2, then \(X_{N}\cong C_{4p}\)  and hence \(\rm {Aut(X_{N})}\cong D_{8p}\) . Since \(K=K_{v}N\)  for any \(v\in V(X)\)  and \(K\)  acts faithfully on \(V(X)\) , we have \(K \leqslant S_{3}\)  and hence \(K_{v}\le 2\) . Then \(|A|\mid 48p \) . Therefore \(P \trianglelefteq A\)  because according to assumption \(p > 53\) . This leads to a contradiction. Now let \(|V(X_{N})|= 4p\)  and \(X_{N}\)  has valency 4. Then \(X_{N}\)  is arc-transitive or half-arc-transitive. By [[9]}, Table 1] and Proposition REF , \(X_{N}\cong C(2;p,2)\) , \(C_{2p}[2K_{1}]\) , \(CA^{0}_{4p}\) , \(CA^{1}_{4p}\)  or \(X(4,p)\) . Let \(X_{N}\cong C(2;p,2)\)  or \(C_{2p}[2K_{1}]\) . Since \(X_{N}\)  has valency 4, \(N\)  acts semiregularly on \(V(X)\)  and so \(X\)  is a \(\mathbb {Z}_{3}\) -regular covering of \(C(2;p,2)\)  or \(C_{2p}[2K_{1}]\) . Assume that \(Y=X_{N}\cong CA^{0}_{4p}\)  or \(CA^{1}_{4p}\)  and \(B= \rm {Aut(Y)}\) . Since \(|K|=|N|\) , we have \(A/K \le B\)  and hence \(|A| \le 48p\) . Then \(P \trianglelefteq A\) . Suppose that \(Y=X_{N}\cong X(4,p)\)  and \(B= \rm {Aut(Y)}\) . Since \(Y\)  is half-arc-transitive, we have \(|B|= 2^{m+2}.p\) , for some nonegative integer \(m\) . Let \(M\)  be a minimal normal subgroup of \(B\) . Thus \(M\)  is an elementary abelian 2- or p-group. First, assume that \(M\)  be a p-group and \(Y_{M}\)  be the quotient graph of \(Y\)  corresponding to the orbits of \(M\)  on \(V(Y)\) . Then \(|V(Y_{M})|= 4\) . Since \(Y\)  is an edge-transitive graph and \(M_{v}=1\)  for \(v \in V(Y)\) , \(Y_{M}\)  has valency 4, a contradiction. Suppose that \(M\)  is a 2-group. Therefore \(|V(Y_{M})|= 2p\)  or \( p\)  and \(Y_{M}\)  has valency 2 or 4.
Subcase 1: \(|V(Y_{M})|= 2p\) .
If \(Y_{M}\)  has valency 2 then \(Y\cong C_{2p}[2K_{1}]\) , which is arc-transitive. Since \(Y\)  is half-arc-transitive, we get a contradiction. Suppose now that \(Y_{M}\)  has valency 4. By [[10]}, Table 1], \(Y_{M}\cong G(2p,4)\)  or \(G(2,p,2)\) . Assume that \(Y_{M}\cong G(2p,4)\) . Since \((K_{1})_{v}=1\) , \(|B/ K_{1}| \leqslant 8p\)  and hence \(|A| \le 48 p\) . It is a contradiction because \(p > 53\) . Suppose that \(Y_{M}\cong G(2,p,2)\) . Let \(Z= Y_{M} \cong G(2,p,2)\)  and \(C= \rm {Aut(Z)}\) . Let \(H\)  be a minimal normal subgroup of \(C\)  and let \(Z_{H}\)  be the quotient graph of \(Z\)  with respect to the orbits of \(H\) . Since \(|C|=2^{p+1}.p\) , \(H\)  is 2- or p-group. Assume that \(H\)  is a 2-group. Thus \(|Z_{H}|=p\)  and \(Z_{H}\)  has valency 2 or 4. By [[4]}, Theorem 3], \(|\rm {Aut(Z_{H})}|=2p\)  or \(4p\) . Assume that \(K_{1}\)  be the kernel of \(C\)  acting on \(V(Z_{H})\) . If \(Z_{H}\)  has valency 4 then \(|K_{1}|=|H|=2\)  because \(|(K_{1})_{v}| =1\) . Then \(C/ K_{1} \leqslant 16p\)  and hence \(2^{p+1}\le 8p\) . We get a contradiction because \(p > 53\) . If \(Z_{H}\)  has valency 2 then \(|K_{1}|\le 8\)  because \(|(K_{1})_{v}| \le 2\) . Thus \(C/ K_{1} \leqslant 16p\)  and hence \(2^{p+1}\le 8p\) , a contradiction can be obtained. Now, suppose that \(H\)  is a p-group. Then \(|Z_{H}|=2 \)  with valency 2, 4, a contradiction.
Subcase 2: \(|V(Y_{M})|= p\) .
If \(Y_{M}\)  has valency 4 then by lemma REF , \(Y\)  is \(\mathbb {Z}_{2}\) -regular cover of \(Y_{M}\)  and \(Y\)  is a Cayley graph. But by [12]}\(, X(4,p)\)  is not a Cayley graph, a contradiction. Suppose that \(Y_{M}\)  has valency 2 and hence \(Y_{M}\cong C_{p}\) . Assume that \(K_{1}\)  is the kernel of \(B\)  acting on \(V(Y_{M})\)  and \((K_{1})_{v} = 1\) . Then \(B/K_{1} \leqslant \rm {Aut(Y_{M})}\)  and so \(|B| \le 8p\) . Therefore \(|A| \le 24p\)  and hence \(P \trianglelefteq A\)  because \(p > 53\) . Then \((K_{1})_{v} \ne 1\) . Let \(V(Y _{M})=\lbrace  \Omega _{0},\Omega _{1},\Omega _{2},..., \Omega _{p-1}\rbrace \) . The subgraph induced by any two adjacent orbits is either a cycle of length 8 or a union of two cycles of length 4. Suppose that \(\langle \Omega _{i}\cup \Omega _{i+1} \rangle \)  is an 8-cycle. Thus \(K_{1}\)  acts faithfully on each \(\Omega _{i}\)  and hence \((K_{1})_{v} \cong \mathbb {Z}_{2}\) . It implies that \(|K_{1}|=8\) . Since \(M\)  is transitive on each \(\Omega _{i}\)  and \((K_{1})_{v} > 1\) , all edges
in the induced subgraph \(\langle \Omega _{i}\cup \Omega _{i+1} \rangle \)  have the same direction either from \(\Omega _{i}\)  to \(\Omega _{i+1}\) 
or from \(\Omega _{i+1}\)  to \(\Omega _{i}\)  in the oriented graph \(D_{B}(Y)\) . It follows that \(B/K_{1} \cong \mathbb {Z}_{p}\)  and \(|B| \le 8p\) . Therefore \(|A| \le 24p\)  and hence \(P \trianglelefteq A\)  because \(p > 53\) . Assume that \(\langle \Omega _{i}\cup \Omega _{i+1} \rangle \)  is a union of two 4-cycles. Let \(\Omega _{i}= \lbrace  u_{i}^{0}, u_{i}^{1}, u_{i}^{2}, u_{i}^{3}\rbrace \)  for any \(i\)  in \(\mathbb {Z}_{p}\) . Then \(B\)  has an automorphism \(\alpha \)  of order \(p\)  such that for any \(i\)  in \(\mathbb {Z}_{p}\) , \(\Omega _{i}^{\alpha }=\Omega _{i+1}\) . Let \((u_{i}^{j})^{\alpha }=u_{i+1}^{j}\)  for \(i\)  in \(\mathbb {Z}_{p}\)  and \(j\)  in \(\mathbb {Z}_{4}\) . Consider a 4-cycle \(C\)  in the induced subgraph \(\langle \Omega _{0}\cup \Omega _{1} \rangle \)  and let \(n\)  be the number of edges of \(C\)  which are in some orbit of \(\alpha \) . Then \(n=0,1\) , or 2. Consequently, the induced subgraph \(\langle \Omega _{0}\cup \Omega _{1} \rangle \)  is one of the of the following three cases.
In the Case 1, \(Y\)  is disconnected, a contradiction. In the Case 2, \(Y\cong C_{2p}[2K_{1}]\) . We get a contradiction because \(Y\cong X(4,p)\) . In the Case 3, \(Y\cong C(2;p,2)\)  that is arc-transitive. It is a contradiction because \(X(4,p)\)  is a half-arc-transitive graph.
<FIGURE>
Fig 3. The induced subgraph \(\langle \Omega _{0}\cup \Omega _{1} \rangle \) .
Case III: \(N\)  is p-group.
If \(|N|=p \)  then \(N\)  is a normal Sylow \(p\) -subgroup of \(A\)  as claimed.
